Proof that $\frac{t \xi_0}{\sqrt{\pi}} + \frac{1}{\sqrt{2\pi}} \sum \frac{\sin(nt)}{n}\xi_n$ is Wiener process. 
Consider $\xi_i$ are i.r.v with standart normal distribution. Let 
  $$X_t = \frac{t}{\sqrt{\pi}}\xi_0 + \frac{1}{\sqrt{2\pi}} \sum_{n=1}^{\infty}\frac{\sin(nt)}{n}\xi_n.$$
We want to prove that $X_t$ is a Wiener process.

My attempt : we need to show that $X_t$ is gaussian, $\mathbb{E}(X_t) = 0$ and $\text{cov}(X_t,X_s) = \min(s,t)$.
First of all if we consider partial sum , we will see that $$\frac{t}{\sqrt{\pi}}\xi_0 + \frac{1}{\sqrt{2\pi}} \sum_{n=1}^{m}\frac{\sin(nt)}{n}\xi_n$$ is gaussian, so the limit will be the same.
Now it's easy to see that $\mathbb{E}(X_t) = 0$.
The main problem in \begin{align*} \text{cov}(X_t,X_s) &= \mathbb{E}(X_t X_s) \\ &= \mathbb{E}\left(\left(\frac{t}{\sqrt{\pi}}\xi_0 + \frac{1}{\sqrt{2\pi}} \sum_{n=1}^{\infty}\frac{\sin(nt)}{n}\xi_n \right)\left(\frac{s}{\sqrt{\pi}}\xi_0 + \frac{1}{\sqrt{2\pi}} \sum_{n=1}^{\infty}\frac{\sin(ns)}{n}\xi_n \right) \right). \end{align*}
If we extract the bracets, we will see that there will be only $$\frac{ts}{\pi} + \frac{1}{2\pi} \sum_{n=1}^{\infty}\frac{\sin(ns)\sin(nt)}{n^2}.$$
The last series I can't evaluate. 
I thought about Fourier decomposition of $\mathbb{1}_{[0,t]}(x)$ because it looks like the same, but calculations give me other result.
Any hints?
 A: I believe that there is a typo in the definition of $X_t$; it should read $$X_t := \frac{t}{\sqrt{\pi}} \xi_0 + \frac{\color{red}{\sqrt{2}}}{\sqrt{\pi}} \sum_{k \geq 1} \frac{\sin(kt)}{k} \xi_k. \tag{1}$$ Moreover, it should be mentioned that this process is only a Brownian motion on the time interval $[0,1]$.

It is known that
$$\varphi_0 := 1 \qquad \varphi_k := \sqrt{2} \cos(k \pi x), \qquad k \geq 1$$
is an orthonormal basis of $L^2([0,1])$. For $r \in[0,1]$ we thus have
\begin{align*} 1_{[0,r]}(x) &= \sum_{k \geq 0} \langle 1_{[0,r]}, \varphi_k \rangle_{L^2} \varphi_k(x)= r + \frac{\sqrt{2}}{\pi} \sum_{k \geq 1} \frac{\sin(k \pi r)}{k} \varphi_k(x). \tag{2} \end{align*}
Now let $s,t \in [0,1]$. Since
\begin{align*} \min\{s,t\} = \langle 1_{[0,t]}, 1_{[0,s]} \rangle_{L^2[0,1]} = \pi \langle 1_{[0,t/\pi]},1_{[0,s/\pi]}\rangle_{L^2[0,1]} \end{align*}
it follows from (2) and the orthogonality of the basis $(\varphi_k)_{k \geq 0}$ that
\begin{align*} &\min\{s,t\} \\ &=  \pi \int_{[0,1]} \left( \frac{t}{\pi} + \frac{\sqrt{2}}{\pi} \sum_{k \geq 1} \frac{\sin(k t)}{k} \varphi_k(x) \right) \cdot \left( \frac{s}{\pi} + \frac{\sqrt{2}}{\pi} \sum_{n \geq 1} \frac{\sin(n s)}{n} \varphi_n(x) \right) \, dx \\ &= \pi  \left( \frac{ts}{\pi^2} + \frac{2}{\pi^2} \sum_{n \geq 1} \frac{\sin(ks) \sin(kt)}{k^2} \right) \end{align*} Following your computations (with the corrected definition of $X_t$, cf. (1)) we conclude that
$$\mathbb{E}(X_s X_t) =  \frac{ts}{\pi} + \frac{2}{\pi} \sum_{n \geq 1} \frac{\sin(ks) \sin(kt)}{k^2}  =\min\{s,t\}.$$
